Is a Markov chain, represented by its transition matrix, always a positive operator?
Let $\mathfrak{A}$ a finite-dimensional $C^*$-algebra. Then it is isomorphic to a direct sum of full matrix-algebras $M_{n_i}(\mathbb{C})$. For simplicity let's consider
$$\mathfrak{A} = M_n(\mathbb{C})$$
Is it true, and why(eventually) that any Markov chain $\Phi : \mathfrak{A} \rightarrow \mathfrak{A}$ (represented by its transition matrix) is a positive operator?(Is it also completely positive?)
EDIT: Consider the 3x3 matrix $$ \begin{bmatrix} a & b & 1-(a+b) \\\\ 0 & 0 & 1 \\\\ 0 & 1 & 0 \\\\ \end{bmatrix} $$
where $0 < a < 1$ and $0 \le b\le 1$.
Is this a positive map from $\mathbb{C}^3 \rightarrow \mathbb{C}^3$?
EDIT2:
It is actually positive, I'm not referring to the matrix itself, which is not positive since $-1$ is an eigenvalue. But it is positive in the sense that it preserves the cone of positive elements of $\mathbb{C}^3$, indeed, taking an element in the positive cone it can be written as $$\begin{bmatrix}|\lambda_1|^2\\ |\lambda_2|^2\\ |\lambda_3|^2 \end{bmatrix}$$
and applying such the matrix transformation to this vector returns an element of this kind (for second and third components is easy).
Basically I was making confusion with the map itself and $C^*$-algebra element
If you thinking of $\mathbb C^3$ as a C$^*$-algebra, then yes. Positive elements in $\mathbb C^3$ are simply triples with non-negative entries. Any matrix with non-negative entries will then map positive elements to positive elements, so your matrix is positive as an operator on the algebra $\mathbb C^3$.
And it is completely positive, because positive maps between abelian C$^*$-algebras area automatically completely positive.